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Lateral Load Distribution of Frame Building
- In a two dimensional moment resisting frame each joint can have at the most three degrees of freedom (displacement in horizontal and vertical directions and rotation).
- Total number of degree of freedom is 3Nj where Nj is the number of joints in the frame.
- In practice, beams carry very small axial force and undergo negligible axial deformation. This means horizontal displacement at all joints located at the beam level s same.
- In most buildings uptown moderate height, the axial deformation of columns is negligible.
- Numbers of degrees of freedom are reduced to one rotation and one horizontal displacement.
- As the rotational inertia associated with the rotational degree of freedom is insignificant, it is further possible to reduce, through static condensation, the number of degrees to one per storey for carrying out dynamic analysis.
- In similar way, each joint of three dimensional frames can have at most six degrees of freedom.
- Finally, there are three degrees of freedom per floor.
- Free vibration analysis of the building can thus be carried out by solving (3N*3N) Eigen value problem, where N is the number of storeys in the building.
- Once natural frequency and more shape is known it is possible to obtain the maximum seismic force to be applied at each storey level due to given earthquake ground motion.
Lateral Load Analysis of Moment Resisting Frame
- Once the design lateral loads are known on the two-dimensional frames, one could analyze the frame for the member forces.
- One could carry out an accurate computer analysis or an approximate analysis as per requirement.
- Approximate analysis is usually performed at preliminary design stage and to assess the computer analysis.
- Two commonly used methods:-
Fig. 1 Portal frame pin-supported at baseFixed-Supported Portals: Portal with two fixed supports, Fig. 2(a) are statically indeterminate to the third degree since there is a total of six unknowns at the supports. If the vertical members have equal lengths and cross-sectional areas the frame will deflect as shown in Fig. 2(b). For this case, we will assume points of inflection occur at the midpoints of all three members, and therefore hinges are placed at these points. The reactions and moment diagrams for each member can therefore be determined by dismembering the frame at the hinges and applying the equations of equilibrium to each of the four parts. The results are shown in Fig. 2(c). Note that, as in the case of the pin-connected portal, the horizontal reactions (shear) at the base of each column are equal. The moment diagram for this frame is indicted in Fig. 2(d).
Fig. 2 Analysis of portal frames - Fixed at basePartially Fixed (at the Bottom) Portal: Since it is both difficult and costly to construct a perfectly fixed support or foundation for a portal frame, it is a conservative and somewhat realistic estimate to assume a slight rotation to occur at the supports, as shown in Fig. 3(a). As a result, the points of inflection on the columns lie somewhere between the case of having a pin-supported portal (as shown in Fig. 1(a)), where the “inflection points” are at the supports (base of columns), and that of a fixed-supported portal (as shown in Fig. 2(a)), where the inflection points lie at the center of the columns. Many engineers arbitrarily define the location at h/3 (Fig. 3(b)), and therefore place hinges at these points, and also at the center of the girder.
Fig. 3 Portal frame partially fixed at baseTrussed Frames: When a portal is used to span large distances, a truss may be used in place of the horizontal girder. Such a structure is used on large bridges and as transverse bents for large auditoriums and mill buildings. A typical example is shown in Fig. 4(a). In all cases, the suspended truss is assumed to be pin connected at its points of attachment to the columns. Furthermore, the truss keeps the columns straight within the region of attachment when the portals are subjected to the sidesway D, Fig. 4(b). Consequently, we can analyze trussed portals using the same assumptions as those used for simple portal frames. For pin-supported columns, assume the horizontal reactions (shear) are equal, as in Fig. 1(c). For fixed-supported columns, assume the horizontal reactions are equal and an inflection point (or hinge) occurs on each column, measured midway between the base of the column and the lowest point of truss member connection to the column. See Fig. 2(c) and Fig. 4(b).
Fig. 4 Trussed frame fixed at base
Lateral loads on Building Frames: Portal Frame MethodThe action of lateral loads on portal frames and found that for a frame fixed supported at its base, points of inflection occur at approximately the center of each girder and column and the columns carry equal shear loads, Fig. 2. A building bent deflects in the same way as a portal frame, Fig.5 (a), and therefore it would be appropriate to assume inflection points occur at the center of the columns and girders. If we consider each bent of the frame to be composed of a series of portals, Fig. 5 (b), then as a further assumption, the interior columns would represent the effect of two portal columns and would therefore carry twice the shear V as the two exterior columns. Consider the 2-D frame with m-base and n-storeys. The degree of indeterminacy of the frame is 3mn. To analyze the frame, 3mn assumptions are made;
- The point of contra-flexure in the column is at mid-height of the columns: (m+1)n assumptions.
- The point of contra-flexure in the beams is at the mid span of the beams: mn assumptions.
- Axial force in the internal columns is zero (m+1)n assumptions.
- With the above assumptions, the frame becomes statically determinate and member forces are obtained simply by considering equilibrium.
- The point of contra-flexure in the beams is at mod span of the beams: mn assumptions.
- Axial force in the columns is approximated by assuming that the frame behaves as a cantilever beam. Neutral axis of the frame is obtained using the column area of cross section and the column location, axial stress in the column is assumed to vary linearly from this neutral axis: (m-1)n assumptions.